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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Jeff Calareso*

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

Use the properties of exponentials and logarithms to learn how carbon dating works. This lesson covers properties of a natural log and rules of logarithms.

Let's return to a scientist studying an old math book. He wants to find out how old it is using carbon dating. He knows that he needs to use an **exponential** to do this. He has *y* = *a*^*x*, where *a* is a **base** and *x* is the **exponent**. You can remember all you need to know about powers and exponentials by thinking back to the bunny problem. You start out with two bunnies, then they multiply. First you have 2, then you have 2 * 2 ... * 2 * 2 * 2. You have 2^*x* bunnies, where you're multiplying 2 times itself *x* times. In the case of the bunnies, 2 was our base.

But there's a special type of base that we see everywhere in the physical world. That base is *e*. *e* is just a constant, like *pi*, except that *e* is Euler's number, and it's roughly 2.72.

Now, *e* follows all of the properties of other powers, like (*e*^*x*)(*e*^*y*) = *e*^(*x* + *y*). *e*^*x* / *e*^*y* = *e*^(*x* - *y*). *e*^0 = 1. (*e*^*n*)^*m* = *e*^(*nm*). In all of these cases, *e* is just 2.72. When I say *e*^0=1, what I'm saying is that 2.72 ... ^0 = 1.

What's special about *e*^*x* is its inverse. *f(x)* = *e*^*x*, but the inverse of *f(x)* is the natural log of *x*: ln(*x*). So *f*^-1(*f(x)*) = ln(*e*^*x*) = *x*. Let's go through some properties of *e* and the natural log. We know that *e*^(ln(*x*)) must be *x*. We know that ln(*e*^*x*) = *x*. But we also know that ln(*x*/*y*) = ln(*x*) - ln(*y*). Similarly, ln(*xy*) = ln(*x*) + ln(*y*). ln(*x*^*n*) = *n*ln(*x*).

So let's use these properties to help date an old math book. Let's use carbon dating. The half-life of a particular type of carbon is 5,730 years. Carbon dating is done by setting the percentage of this special type of carbon as being equal to *e* to some constant number times *t* (*e*^(*Ct*)).

If you set the percentage of carbon that you have, and you know what this *C* is, you can calculate how old your item is. First, we need to find out what this number *C* is. I said that the half-life of carbon is 5,730 years. That means that after 5,730 years, we have 50% of our carbon left. So 50% carbon = *e*^(*C*(5730)). Okay, well, 50% carbon is the same as .5, so I have the equation .5 = *e*^(5730*C*). To solve for *C*, I need to first take the natural log of both the left-hand side and the right-hand side. I get ln(.5) = ln(*e*^(5730*C*)). If I use my property that ln(*e*^*x*) = *x*, the right-hand side is equal to 5730*C*. If I divide both the left and right side by 5730, I can calculate *C*. *C* = ln(.5)/5730, which is about equal to -0.000121. So let's put that back into our original equation. The percent carbon that we have is equal to *e*^(-0.000121*t*). So if I plug in the percent carbon that I have on the left-hand side, I can solve this equation for the age of my material.

Say that there's 30% of this special carbon in our math book. I'm going to take the natural log of both the left-hand side and the right-hand side. I get ln(.3) = ln(*e*^(-0.000121*t*)). I know that ln(*e*^*x*) = *x*, so the right-hand side simplifies to -0.000121*t*. I can divide both sides by that number, and I find that *t* = (ln(.3))/-0.000121, which is about 9,950 years old. That's a really old math book. While we see *e* and the natural log a lot in the physical world, there are other bases that have similar properties.

Before we had *y* = *e*^*x*, where *e* was the base. We can also have *y* = 10^*x*, where 10 is now our base. When *y* = *e*^*x*, the inverse is the natural log. So in the case of *y* = *e*^*x*, the natural log of *y* = *x*. If we have a different base, we write this not as natural log, but as log with some base. So if we have *y* = 10^*x*, then the inverse is log base 10. So *y* = 10^*x* is the same thing as saying log base 10(*y*) = *x*.

There's an important connection here. The natural log is the exact same thing as writing log base *e*. But it's so common, we just call it the natural log. As with the case for natural logs, log base 10(10^*x*) = *x*. log base *b*, where *b* is any number, of *b* equals 1. log base *b*(1) = 0; that's like saying the natural log of 1 is equal to 0. The natural log of *e* is equal to 1. All of these properties are the same, regardless of whether you're looking at the natural log or the log.

So let's review. If you have an equation like *y* = *a*^*x*, we call *a* the **base** and *x* the **exponent**. The inverse of this is log base *a*(*y*) = *x*. If the base is *e*, log base *e* is the same thing as saying the natural log.

Both exponents and logs have similar properties. (*e*^*x*)(*e*^*y*) = *e*^(*x* + *y*). *e*^*x* / *e*^*y* = *e*^(*x* - *y*). And *e*^0 = 1. (*e*^*n*)^*m* = *e*^(*nm*).

On the log side, ln(*xy*) = ln(*x*) + ln(*y*). ln(*x*/*y*) = ln(*x*) - ln(*y*).

Similarly, ln(*xy*) = ln*x* + ln*y*. ln(*x*^*n*) = *n*ln(*x*). ln(1) = 0 and ln(*x*^*n*) = *n*ln(*x*). All of these are true because *e*^(ln*x*) = *x* and ln(*e*^*x*) = *x*.

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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

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